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# 'Faster than light' question

1. ### John_M

24
According to Einstein nothing can move faster than c. But whereas c is constant in all frames of reference, the speed of massive objects is defined relative to other massive objects. So when we say nothing can go faster than c, we surely mean nothing can go faster than c relative to another object.

So - I've heard of particles that travel quite close to c relative to the Earth. If I fire a particle like this in one direction, it will be travelling at a speed greater than 0.5 c relative to the Earth. If I then fire another particle of the same speed in the opposite direction, it will be travelling greater than 0.5c relative to the Earth and greater than c relative to the other particle. Correct?

Is it then correct to state that, in one of these particles' frame of reference, the other particle will be going backwards in time?

2. ### chroot

10,427
Staff Emeritus
From your Earthly perspective, one particle will be moving at 0.5c in one direction, and the other will be travelling at 0.5c in the other direction.

From the particles' perspective, the earth is moving at 0.5c, but the other particle is moving at

$$\frac{0.5c + 0.5c}{1 + \frac{(0.5c)^2}{c^2}} = 0.8c$$

Velocities in special relativity do not add in the normal Galilean way.

- Warren

699
From an observer on earth, particles can be moving apart at 2c, but as long as this earthly observer is stationary and not in the motion frame of the paticles. Relative to the particle, you need to use relativistic velocity addition, like chroot said.

4. ### John_M

24
How do they add then? It's news to me

I read something about 'neutral pions' going at 185,000 miles per second - presumably relative to the Earth - in experiments to test the speed of light. How does your statement that 'From your Earthly perspective, one particle will be moving at 0.5c in one direction, and the other will be travelling at 0.5c in the other direction' square with this? If you accept that it's possible for a particle to travel at a speed greater than 0.5c relative to the Earth (and it does seem to be) it seems odd to suggest that this particle will slow down to 0.5c relative to the Earth simply because another particle has been fired in the opposite direction. This is my problem (although I'm sure that's not what you're actually suggesting...)

Thanks

John.

5. ### John_M

24
actually, I could have read over your post a bit more closely...you've posted an equation for working out relativistic velocities...

6. ### Janus

2,374
Staff Emeritus
No, this is not what he is suggesting. Both particles maintain a velocity of .05c relative to the Earth as measured by the Earth. Each particle also measures its relative velocity to the Earth as 0.5c. Each particle, however will measure the others particles relative velocity to itself as 0.8c (not 1c). if the particles relative velocity to the Earth were 0.75c, they would measure their respective relative velocity as 0.96c

7. ### John_M

24
"Both particles maintain a velocity of .05c relative to the Earth as measured by the Earth. Each particle also measures its relative velocity to the Earth as 0.5c."

It's this statement I don't understand. Do you accept that it's possible for a particle to travel at greater than 0.5c relative to the Earth and, if so, what's your basis for stating that 'both particles maintain a velocity of 0.5c relative to the Earth'? As I understand it, some particles are capable of travelling at 0.99c relative to the Earth - so where does the figure of 0.5c come from?

Thanks for the replies

John.

8. ### chroot

10,427
Staff Emeritus
You're the one who brought up 0.5c. Bottom line is that no observer will ever see another object going faster than light.

An observer might see one object leaving at almost the speed of light in one direction, and another object leaving at almost the speed of light in another direction, but that doesn't violate that bottom line.

- Warren

9. ### John_M

24
"An observer might see one object leaving at almost the speed of light in one direction, and another object leaving at almost the speed of light in another direction, but that doesn't violate that bottom line."

Why not? It seems very clearly to violate the bottom line to me, seeing as relative speed is the only type of speed that exists in SR and there's no reason to prioritise the observer's frame over that of the objects. If what you say is true, doesn't it follow (if the objects are moving in directly opposite directions) that the first object will be moving at almost 2c relative to the second?

Tell me where I've gone wrong here. Assume all speeds are constant.

1) I fire a particle at a speed of 0.99c relative to the Earth.
2) I fire a second particle at a speed of 0.99c relative to the Earth in the opposite direction to the first particle.
3) Therefore Particle 1 is travelling at a speed of 1.98c relative to Particle 2.

Now, you might state that it's impossible to go faster than c, full stop, mass increases will happen etc etc. But saying 'it's impossible to go faster than c' - doesn't tell the whole story because it doesn't specify a frame.

If you accept that I can, at will, fire a particle at 0.99c - and in my limited understanding scientists have done just that - e.g. 'neutral pions' used to test the constancy of the speed of light - what exactly is stopping me from firing a second such particle in the opposite direction, and isn't it the case that the second particle will have a speed of 1.98c relative to the first?

10. ### chroot

10,427
Staff Emeritus
Again, you're mixing reference frames. You can't assert that an observer on a moving particle will measure another moving particle the same way as a stationary observer between them.

If you fire two particles at 0.99 c from earth in opposite directions, here are the relative speeds seen by all observers:

Particle 1:
Sees Earth moving at 0.99c
Sees Particle 2 moving at 0.999949498c

Earth:
Sees Particle 1 moving at 0.99c
Sees Particle 2 moving at 0.99c

Particle 2:
Sees Earth moving at 0.99c
Sees Particle 1 moving at 0.999949498c.

In no case is any speed > c.

- Warren

11. ### John_M

24
Where do you get the number 0.999949498c from?

12. ### chroot

10,427
Staff Emeritus
13. ### jcsd

2,226
John -

You're applying a Gallielan transformation, but what special relativty says is: "the Gallilean transformation is only an approximation that applies when the velcoites involved are much smaller than c". What you should be applying is a Lorentz transformation.

14. ### John_M

24
OK...thanks guys.

I think the statement from Chroot at the start clears it up...

'Velocities in special relativity do not add in the normal Galilean way.'

699
yes, the observer on earth can see the two particles moving away from eachother at 0.99c + 0.99c.

16. ### LURCH

2,512
John,
If you're looking for a more intuitive understanding, this is where time dilation comes in. Partical 1 sees Partical 2's movement through time as greatly slowed down. So Partical 2 may measure his speed as .99c relative to the Earth, but Partical 1 would say, "well sure he measures his own speed as 186,000 miles per second, but look at his watch; his seconds are almost twice as long as they should be!"

17. ### robphy

4,354
It is the "rapidities" (analogous to the Euclidean angles) that add.
Velocity is the hyperbolic-tangent of the rapidity.
The "addition [better: composition] of velocities" formula is a trigonometric identity involving the hyperbolic-tangent of the sum of two rapidities.

18. ### jcsd

2,226
And once you look the Lorentz transformation in terms of the rapidity, the simlairty between boost (a change in velocity) and rotation become obvious.

19. ### robphy

4,354
On top of that, calculations are much easier to execute and interpret because one can appeal to one's intuition on [hyperbolic-]trigonometric and exponential functions.

For the uninitiated,
$\gamma$, $\gamma v$, $v$ are $\cosh(\theta)$, $\sinh(\theta)$, and $\tanh(\theta)$.

Exercise: find the physical interpretation of $\exp(\theta)$. Hint: express $\exp(\theta)$ in terms of $v$.

Last edited: Aug 20, 2004
20. ### jcsd

2,226
$$e^{\theta} = \cosh\theta + \sinh\theta = \gamma + \gamma v=$$
$$\gamma(1 + v) = \frac{1 + v}{\sqrt{1 - v^2}} =$$
$$\sqrt{\frac{(1 + v)^2}{1 - v^2}} = \sqrt{\frac{1 + v}{1 - v}} = \frac{x + t}{x' + t'}$$

latex is on the blink, the last terms should be (x + t)/(x' + t')

Last edited: Aug 20, 2004