# Speed of Cars while running toward them

• July 16th 2008, 04:52 PM
Jonboy
Speed of Cars while running toward them
If a car is coming directly toward me going 35 mph, and I am running 6 mph towards it, what speed from the observer in the car will it look like I am going?

This is not homework. I am reading a book on Einstein and his theory of relativity has caused me to think of this.
• July 16th 2008, 05:08 PM
mr fantastic
Quote:

Originally Posted by Jonboy
If a car is coming directly toward me going 35 mph, and I am running 6 mph towards it, what speed from the observer in the car will it look like I am going?

This is not homework. I am reading a book on Einstein and his theory of relativity has caused me to think of this.

Taking a purely Newtonian approach (the approximation is valid for these low speeds): 41 mph.
• July 16th 2008, 05:11 PM
Jonboy
Huh? Wouldn't I look even slower because someone is driving toward me about 6 six times faster than me?
• July 16th 2008, 05:14 PM
topsquark
Quote:

Originally Posted by Jonboy
If a car is coming directly toward me going 35 mph, and I am running 6 mph towards it, what speed from the observer in the car will it look like I am going?

This is not homework. I am reading a book on Einstein and his theory of relativity has caused me to think of this.

At those speeds? 35 mph + 6 mph = 41 mph.

But if you really want to apply relativity to this then set yourself as an inertial reference frame. The ground, the "moving frame" in this problem, is going 6 mph, so this is the v you are going to use for $\gamma$. (Though we don't actually need to calculate $\gamma$ for this.) The truck's speed is 35 mph with respect to the ground. The truck is coming at you so the truck appears to be moving at
$\frac{6 + 35}{1 - \frac{35 \cdot 6}{c^2}}$
where c is in mph.

I'm not going to bother to calculate it, but the result is only infinitesimally different from 41 mph.

-Dan

Edit: I got it backward. I did the problem of how fast the truck would look according to you. However the answer is still the same. Note though, $\gamma$ should be calculated using the 35 mph speed.
• July 16th 2008, 05:14 PM
mr fantastic
Quote:

Originally Posted by Jonboy
Huh? Wouldn't I look even slower because someone is driving toward me about 6 six times faster than me?

The frame of reference is the driver. You want to know how fast the driver thinks you're going. Note that the driver thinks that s/he is NOT moving ....
• July 16th 2008, 05:24 PM
Jonboy
Quote:

Originally Posted by mr fantastic
The frame of reference is the driver. You want to know how fast the driver thinks you're going. Note that the driver thinks that s/he is NOT moving ....

Why would s/he think s/he is not moving? That sounds illogical. This is merely a humble thought, I am not trying to be a rebel.
• July 16th 2008, 05:24 PM
Jonboy
Quote:

However the answer is still the same.
I don't see how.
• July 16th 2008, 05:43 PM
mr fantastic
Quote:

Originally Posted by Jonboy
Why would s/he think s/he is not moving? That sounds illogical. This is merely a humble thought, I am not trying to be a rebel.

Because they are moving at a constant speed and the calculations are being performed in their rference frame.

You need to go back a few pages and read the stuff about frame of reference, Galilean transformations etc. because at the moment you don't have the conceptual understanding of these things that's required to understand the answer to your question.
• July 16th 2008, 06:42 PM
ticbol
Quote:

Originally Posted by Jonboy
Why would s/he think s/he is not moving? That sounds illogical. This is merely a humble thought, I am not trying to be a rebel.

Okay, in reality, the other driver sees and feels himself as moving at 35 mph towards you.

Suppose he thinks he is not moving? Then it will appear to him that the road is coming to him at the rate of 35 mph.
Now you are riding on that same approaching road with an additional 6 mph relative to the moving road.
Hence the other driver sees you approaching him at (35 +6) = 41 mph.
• July 17th 2008, 06:30 AM
topsquark
Quote:

Originally Posted by Jonboy
Why would s/he think s/he is not moving? That sounds illogical. This is merely a humble thought, I am not trying to be a rebel.

Obviously if we are on the surface of the Earth, or have some other reference that we take as "non-moving" we can tell if we are moving or not. But consider if this were taking place in otherwise empty space. If you were moving with a constant velocity you would have no possible way of knowing that you were moving at all. Take this one step further. If you are standing stationary with respect to the Earth are you moving? Of course. You are moving along with the motion of the Earth. Do you feel like you are moving? Do you use that motion to predict how different objects are going to react in a Physical experiment? Of course not. Most argue that we can ignore it because the Earth is not moving very fast. But it is. The point is that it is not accelerating very fast and it is the acceleration of an object that is what tells us if our reference frame is moving or not. (However Einstein took even that away in General Relativity. I'm not going to discuss that here.)

One of the main points of use with regard to Special Relativity is that the observer is considered to be stationary and all observed objects are measured with respect to that frame. This occasionally causes apparent logical inconsistencies, as in this case, but is essential to the application of the theory.

-Dan
• July 17th 2008, 06:39 AM
topsquark
Quote:

Originally Posted by Jonboy
I don't see how.

This is in reference to my comment that I took the wrong observer as my stationary frame, but that the speed of the observed motion would be the same for the other observer.

The main tenet behind Special Relativity is that two observers in different inertial reference frames measure exactly the some properties of motion. So if one observer measures that a truck is moving toward him at a given speed then the driver of the truck must also measure the observer moving toward him at the same speed.

This can be easily verified in this simple example by the equation used to add velocities:
$\frac{v + u}{1 - \frac{uv}{c^2}}$
When you switch the roles of v and u in the equation the "sum" of the velocities is still the same.

If you are wondering why this principle doesn't work for the "Twin Paradox" the answer is that one twin is accelerating with respect to the other twin at four different times, so we can indeed specify which twin is in motion. Thus there will be differences in the properties of motion of the two twins. Specifically, since an acceleration is involved this problem is a matter for General Relativity. Curiously I have never seen an analysis of this problem done using GR.

-Dan
• July 17th 2008, 07:05 AM
mr fantastic
Quote:

Originally Posted by topsquark
[snip]
If you are wondering why this principle doesn't work for the "Twin Paradox" the answer is that one twin is accelerating with respect to the other twin at four different times, so we can indeed specify which twin is in motion. Thus there will be differences in the properties of motion of the two twins. Specifically, since an acceleration is involved this problem is a matter for General Relativity. Curiously I have never seen an analysis of this problem done using GR.

-Dan

I was sure I'd seen the analysis done in Misner and Thorne but when I checked my copy it appears only as an exercise. There's a tick and an exclamation mark pencilled next to it in my copy so I guess I must have done it at some stage. Someday I might rummage through my archives (unless the silver fish have got to them).
• July 17th 2008, 07:13 AM
mr fantastic
Quote:

Originally Posted by topsquark
[snip]
If you are wondering why this principle doesn't work for the "Twin Paradox" the answer is that one twin is accelerating with respect to the other twin at four different times, so we can indeed specify which twin is in motion. Thus there will be differences in the properties of motion of the two twins. Specifically, since an acceleration is involved this problem is a matter for General Relativity. Curiously I have never seen an analysis of this problem done using GR.

-Dan

Thinking about it some more, I have a feeling the analysis might be done in Lightman et al (Problem Book in Relativity and Gravitation).
• July 17th 2008, 11:05 AM
Jonboy
Thank you all for you input! It is very important to me.
• July 19th 2008, 11:59 AM
mathwimp
Quote:

Originally Posted by topsquark
This is in reference to my comment that I took the wrong observer as my stationary frame, but that the speed of the observed motion would be the same for the other observer.

The main tenet behind Special Relativity is that two observers in different inertial reference frames measure exactly the some properties of motion. So if one observer measures that a truck is moving toward him at a given speed then the driver of the truck must also measure the observer moving toward him at the same speed.

This can be easily verified in this simple example by the equation used to add velocities:
$\frac{v + u}{1 - \frac{uv}{c^2}}$
When you switch the roles of v and u in the equation the "sum" of the velocities is still the same.

If you are wondering why this principle doesn't work for the "Twin Paradox" the answer is that one twin is accelerating with respect to the other twin at four different times, so we can indeed specify which twin is in motion. Thus there will be differences in the properties of motion of the two twins. Specifically, since an acceleration is involved this problem is a matter for General Relativity. Curiously I have never seen an analysis of this problem done using GR.

-Dan

Firstly I can't figure out why, well, not just you Dan, but others too, just didn't do the simple workings out, and give Jonboy the straight numerical answer, which is:

-41.0000000000 mph on a TI 85 calculator!!

but is:

40.999999999999980855037744606653 mph on Windows XP calculator!!
(with a lot more undisplayed places behind that)

The latter btw, seems to have gotten astonishingly accurate since the huge snafoo Intel made with their ALU - I'm not just going off this - but I know from experience that when it calculates an "awkward" proper fraction (specifically 34/109) and shows the recurrence gap of 108 places behind the decimal, there's something pretty good about it ;-)

More seriously:

You either need two plus signs in the velocity composition formula you gave. or two minus signs, at least according to the original derivation. Thus:

$u = \frac{u' + v}{1 + \frac{u'v}{c^2}}$ ... (1)

... which means that an object moving with velocity $u'$ measured in a "primed" reference frame which frame itself is moving with velocity $v$ relative to the unprimed frame will be measured in the unprimed frame as moving with velocity $u$.

The inverse of this gives the two minus signs:

$u' = \frac{u - v}{1 - \frac{uv}{c^2}}$ ... (2)

... giving the velocity $u'$ as measured by "prime" in terms of the velocity $u$ measured by "unprime", where now unprime is regarded as moving with velocity $-v$ relative to prime.

Unfortuneately implicit and not explicit in the original derivation is that v regarded as standing on its own is always to be read as |v|, ie: positive. This implicitness is contained in the phrase: "The inverse formula(e) are obtained by replacing v by -v". The formula as you give it will work with really careful attention to the signs, and then the opposite of the inverse formula you give goes to:

$u = \frac{u' - v}{1 + \frac{uv}{c^2}}$

... so that in any case the +/- signs have to be reversed for the other observer (reference frame)! Hardly simpler. I've got to admit I'm out of touch with current practice (foibles), and that might be the modern way.

If you get the sign on the denominator wrong, you destroy causality - the effects of things can happen before the things themselves!

Anyway so that Jonboy can see how I worked his sum out, I used eq. (2) above, set u = -35 mph (assuming moving along unprimed's -ve x direction), v = |v| = 6 mph, so that u - v = -35 - |v| = -35 - 6 = -41; and 1 - uv/c² = 1 - (-35)|6|/670,616,629.384² to give the answers above - unresolvable from (-)41 mph on my TI-85, which would have to have 10,000 times more resolution to give a more accurate answer!

Numbers that the TI-85 can handle: if u (u') and v are opposite, and each exactly half the speed of light, Newton would say the observed relative velocity would be the speed of light; disallowed by Einstein, who says, according to the formula, that the observed relative velocity would be c/(1 + 0.25) or only 0.8c.

A more obvious example, a "clincher" really, is that if u (u') and v are each just a snot less than c, then the formula gives the velocity between them as observed by a 3rd party "on the ground" as: 2c/(1 + 1) = c; showing c to be the absolute upper limit for any relative velocity whatever the u (u') (<c) and v (<c). Newton, of course, says the relative velocity would be 2c.

The extrovert travelling twin doesn't have to accelerate relative to her boring stay-at-home drone Martha Stewart sister four times; 3 times will do the job: outgoing acceleration, 'turn-around" (one acceleration) and stopping on return.

In fact if, hypothetically of course, the traveller could catch into orbit of some distant black hole, and tangentially come close to her sister each orbital period, well, it's tricky to say how many accelerations are involved there: I'm tempted to say just two - one to get into orbit; but of course, the acceleration in orbit is absolutely continuous, and a "reverse" acceleration to get out; to return to Martha's company again, and probable continued alcoholic depression.

I was wondering about some of this stuff quite some time ago now, went on some forums (fora?) and so on, and came to the conclusion that the twin "paradox" explanation is entirely General Relativity, shouldn't be too shocking as SR is just a less accurate sub-set of GR, at least as far as my understanding goes.

I set this webpage up in relation to that; some of it's probably dog-doo ;-), as admitted there, but I think there's something to it - in particular the planet radius SR-GR time dilation effects "cancelling" orbit at R/2 above a planet's surface:

Twin Paradox, Freely Falling Satellites, General & Special Theories

I wasn't sure whether to be pleased or dissapointed when I managed to hunt down this confirmation of my meanderings on the webpage:

Intersecting Orbits

I was pleased that it confirmed my feelings about GR subsuming SR, and the articificiallity of the bifurcation of a problem into an "SR" part and a GR part, and so that I wasn't going completely mad.

The confirmation about the R/2 orbit (referred to as 3R/2 in the last link) I was more ambivalent about. On the one hand I was pleased to get confirmation, whereas on the other I was obviously not the first (apparently merely the most impressed), to have stumbled upon it. Then again, if I had not found that confirmation, I would have had trouble anyway believing in the work that I'd done! ... well without grinding through the GR stuff myself ... and I think I already said, I'm really rusty ...

And, of course, confirmation (of R/2) that I wasn't well on my way to bonkerdom was again reassuring. ;-)

Really I'd done all that stuff on that webpage just to answer some stuff on a long ago forum, and sort some things out in my own head.

It's still a mystery to me why the R/2 thing impressed me so much, and I hadn't heard of it before discovering it - (not bragging - I did discover it, just because it's all GR old hat doesn't change that - I didn't know about that, and in any case came to the result by a different method), and yet it seems to be buried so deep and unremarked? For some reason others don't seem to share my wonder at that result?

When that result came out so simple, you could have knocked me over with a feather. I thought it was really cool (man ;-). I wonder if anyone else here thinks that the R/2 thing is "really cool"* too?

;-)

Anyways, back to real life, I have this problem with this paraboloid ...

What would it look like. travelling at c, bullet shaped end first as it passed a (lucky) observer?

No, that's not it. THIS IS IT.

Dennis R.

*No, I'm not an ex-hippy.
(not that there's anything wrong with that)