# Understanding the lightspeed limit

1. Feb 16, 2014

### daveyb

I'm trying to understand the special theory of relativity. All answers I've found to the problem below seem to provide an inadequate answer. Could anyone answer this question?

A rock in space at position B explodes into two halves (A & C) in such a (hypothetical) way that each half moves away from position B at a constant speed of 0.6 x the speed of light. Regardless of what anyone in any position might observe, how fast is A moving away from C?

Is it 1.2c, and if not, aside from the fact that we're simply told that an object cannot move faster than the speed of light, why can't A be moving away from C at 1.2c? or can it?

Thanks for any thoughts

N

2. Feb 16, 2014

### BruceW

hi, welcome to physicsforums :)
yeah, as you suspect, it is not possible for them to explode away from each other at relative velocity of 1.2c. A simple way to realize why this is impossible is to think of the case when they explode away from each other at some relative velocity very close to c, yet still less than c. OK, so thinking about the observer looking at the stationary rock at position B: If he puts some dynamite inside the rock, then this energy released by the explosion will be converted into the kinetic energy of the two halves as they fly away from each other, right? And so here is the critical bit: the more dynamite you use, the more kinetic energy the two halves will have, right? So you might intuitively think that you can add so much dynamite that the two halves will fly away at greater than c. But, this is not the case, because it takes an infinite amount of energy to make the two halves fly away from each other at c. More precisely, you could keep adding more energy to your dynamite pile, and therefore the two halves would have arbitrarily large kinetic energy. But still, their relative velocity will be less than c.

In short, no matter how much dynamite you use, and how much kinetic energy you give to the two halves, this will still correspond to a relative velocity which is less than c.

3. Feb 16, 2014

### daveyb

Blazar jets problem

Many thanks for your reply. It somewhat seems to makes sense that you would need an ever increasing amount of energy to explode two objects away from each other as you got them to move close to the speed of light relative to each other. However, what confuses me is that it seems feasible to be able to move one object away from another at close to the speed of light, and if so, why couldn't you do this twice with independent pieces of matter moving in opposite directions. Take the following (potentially) more realistic example:

In reference to http://www.space.com/694-blazing-speed-fastest-stuff-universe.html

A blazar gas jet (Jet A) shoots out from a blazar at close to the speed of light (0.999c). A fixed distance away from the first blazar, a second blazar shoots out a blazar gas jet (Jet B) in the opposite direction at close to the speed of light (0.999c). These events happen simultaneously in the universe in the sense that if an observer was mid way between the jets then light from the tip of the jets would reach him/her at the same time.

Although someone couldn't witness or measure the two jets moving away from each other at close to twice the speed of light (due to the time it takes the light from each to reach a common measuring device), could we not deduce that given we know how far apart the blazars are, how fast blazar gas jets move, and which direction they were pointing in, that Jet A moved away from Jet B at close to twice the speed of light?

If not, given the assumption that it is possible for two blazar jets to exist in the universe at the same time, pointing in opposite directions, and that blazar jets can move at close to the speed of light, how do we conclude that they aren't moving away from each other any faster than the speed of light? or are they?

Last edited: Feb 16, 2014
4. Feb 16, 2014

### Bill_K

In relativity, we use a special formula for combining two velocities - you don't just add them together! This formula guarantees that the result will always come out less than c.

5. Feb 16, 2014

### Staff: Mentor

If you are "stationary" at point C, in your reference frame the two jets are separating at a rate of close to 2c. Relativity has no problem with this, because this is not the velocity of any object in any reference frame.

If you are "riding along" with either jet A or jet B, in that reference frame the other jet is moving with a velocity whose magnitude is very slightly less than c, according to the relativistic "velocity addition" formula that Bill K pointed you to.

6. Feb 16, 2014

### BruceW

yeah, it is counter-intuitive to most everyday life. We do not naturally think in a 'relativistic' way, in my opinion. The answer is essentially what Bill_K said. If you send one object in one direction at speed 0.9c relative to you, and you send another object at speed 0.9c relative to you (in the other direction), this does not mean the relative velocity between the objects is greater than c. In relativity, you can't just add or subtract velocities in the usual (Galilean) way.

7. Feb 16, 2014

### daveyb

Thanks for your answer. I was aware of Lorentz transformation, but thought it was based on the difference between what one observer might see compared to another and is the result of how long it takes the light from the objects to reach an observer.

I know that due to the fixed speed of light, that Rock A could never observe Rock C moving away from it at 1.2c or vice-versa. However, regardless of the limitations of our light-based observation, surely the rocks are actually moving away from each other at 1.2c.

What if when the rocks divide we started an atomic countdown clock on each and made it so that each rock automatically stopped exactly after 1 year (using each rock's own measure of time). Then after that year we walked from one rock to the other (we'd better take some sarnies), wouldn't we find that they were 1.2 light years apart even though they had only travelled away from each other for 1 year suggesting they travelled at a relative speed of faster than light speed?

Last edited: Feb 16, 2014
8. Feb 16, 2014

### A.T.

It has nothing to do with limitations of our light-based observation. Lorentz transformation is what is left after signal delay has been accounted for.

Nothing moves at >c relative in an inertial frame.

If the rocks accelerate, they are not inertial any more. Things can move at any speed in non-inertial frames.

9. Feb 16, 2014

### daveyb

I accept this, but I guess what I'm looking for, which I'm struggling to find anywhere in clarity, is the reasoning behind these postulates other than that Lorentz's formula never outputs a number greater than 1. I'm trying to understand the reasoning behind the formula. How it is so.

Sorry to be frustrating - just trying to understand.

Thanks for everyone's kind replies

10. Feb 16, 2014

### A.T.

They fit experimental evidence.

11. Feb 16, 2014

### Staff: Mentor

Right, their relative speed will be less than c.

I don't know what you mean by "actually". As jtbell explained, there's no problem for a third frame (the rest frame of B, which sees each rock move at 0.6c) to measure the rate at which the rocks separate--according to his frame--to be 1.2c.

12. Feb 16, 2014

### daveyb

I did send thanks to jtbell, but what prevented resolve for me was being told that the time delay (due to how long it might take light to travel from the object to an observer) is taken into account in Lorentz transformation.

By "actually" I mean how fast the molecules in Rock A are travelling relative to those molecules in Rock B as opposed to the observation of them based on measuring the light reflecting off (or emitting from) them.

In my example with the sarnies, would I actually measure a distance of 1.2 light years or would I measure less than 1 due to some distortion in space time?

I obviously have more studying to do!

13. Feb 16, 2014

### BruceW

ah, Doc Al, you beat me to it. I was just about to say the same thing. I guess I could also add - In Euclidean space, the rate at which the rocks separate, according to observer B, is the same as the rate at which the rocks separate according to an observer moving with one of the rocks, or moving with any other velocity. But in Minkowski spacetime, the rate at which the rocks separate according to observer B is not the same rate at which the rocks separate according to an observer moving with one of the rocks.

14. Feb 16, 2014

### Staff: Mentor

Whenever you make a measurement using light you must take into account the time it takes for light to reach you. Otherwise you'll get silly results. But nothing in the Lorentz transformations requires you to use light.

Again it depends on whose frame you would like to consider. In physics, the relative velocity of A with respect to B means the speed of A as measured in a frame in which B is at rest. That relative velocity will always be less than the speed of light.

Observers in frame B will measure the rocks to be 1.2 light years apart after 1 year has passed. According to them! (Other frames will disagree, of course.)

15. Feb 16, 2014

### dauto

The correct addition formula is $$v_{rel}=\frac{v_1+v_2}{1+\frac{v_1v_2}{c^2}},$$ Plugging in the values $$v_1=0.6c,$$ and $$v_2=0.6c,$$ we get $$v_{rel}=\frac{0.6c+0.6c}{1+\frac{(0.6c)(0.6c)}{c^2}}=\frac{1.2c}{1+0.36}=\frac{1.2}{1.36}c=\frac{15}{17}c.$$ Clearly that is still less than the speed of light.

16. Feb 16, 2014

### BruceW

I think daveyb is talking about the separation of the two rocks, according to a guy just standing in the middle. In which case, the rate of increase of separation of the two rocks is faster than c.

17. Feb 16, 2014

### daveyb

Thanks everyone, I think I need to spend more time understanding the experiments behind it all and their outcomes.

18. Feb 16, 2014

### dauto

Yes, that's right. And he is missing the fact that their rate of separation (as measured by a 3rd party) is not a measure of their relative speed (as measured by either one of them)

19. Feb 16, 2014

### Staff: Mentor

In the usual approach to special relativity everything is based on two postulates:
1) the laws of physics are the same in all inertial frames
2) the speed c is the same in all inertial frames

From those two postulates you can derive the Lorentz transform and from the Lorentz transform you can derive all of the other special relativistic effects.

20. Feb 16, 2014

### curious bishal

I am amazed why we should use Euclidean transformation of spacetime and Lorrentz transformation of Spacetime for this problem. In what sense it is different than measuring relative velocities in case of earth. There is a frame of reference i.e B (in rest) like earth while we do some experiment on the earth.
Suppose two cars are moving away from each other here and should we have to use such transformations for their relative velocities? Isn't this the same case in sense that there is also a point of reference which is in rest i.e. B????

21. Feb 17, 2014

### Staff: Mentor

In the case of two cars moving away from each other on earth, the difference between $u+v$ and $\frac{u+v}{1+uv}$ is so small that we neither notice nor care, so use the simpler formula. It's similar to the way that we don't correct for the curvature of the earth when we're laying out the foundations of a house, just act as if the earth is flat.

[edit: you'll notice that I wrote the formula as $$\frac{u+v}{1+uv}$$ instead of $$\frac{u+v}{1+\frac{uv}{c^2}}$$. They're the same as long as $c=1$, which it will be if you measure distances in light-seconds and time in seconds, so that the speed of light comes put to be one light-second per second. Using units in which $c=1$ doesn't change the physics, but it simplifies the formulas a bit]

Last edited: Feb 17, 2014
22. Feb 17, 2014

### BruceW

I'm not totally sure what the question is. But the 'true' relativistic answer is that if each car is moving away from $B$ at speed $v$ (according to $B$) in opposite directions, then the rate of increase of separation of the cars (according to $B$) is $2v$. But, according to someone in one of the cars, the rate of separation of the cars will be some value less than $2v$ (let's call it $2v-w$). As Nugatory says, if $v$ is small compared to $c$, then $w$ is very close to zero, so we don't usually notice these kinds of effects in our everyday lives. However, at the other extreme, if $v$ is very close to $c$ then $w$ must be almost $v$ itself. This is because according to $B$, the two cars are separating at rate almost $2c$, but according to a person inside one of the cars, the two cars are separating at rate almost $c$ (since this is the upper speed limit), so the difference between the velocity according to the two people must be almost $c$.