# Faster than light illusion

1. Feb 17, 2009

### Lexus Dominus

Hello, im really curious about special relativity but dont have a strong background in maths or theoretical physics. would be very greatful if you could help me with this scenario;

XYZ are objects in space

Y explodes violently, sending X and Z away from Y in opposite directions at C / 2 + 1 m/s

is this scenario possible within the realms of SR? if yes, would light emitted from X ever reach Z?

2. Feb 17, 2009

### mgb_phys

If you mean X-Z are moving in opposite directions then yes they can move apart faster than light and will not be able to communicate with each other again - ie light from X can't catch-up with Z. (assuming a flat or expanding universe)

If you mean X,Y,Z in the normal grid coordinates sense, so X is going right and Z is going up, then the distance between them increases as sqrt(2) = 1.4 * the distance along the x (or Z) axis and so the ligth could reach from one to the other.

You don't need an explosion in spcae to do this, the electrons in your TV are going at 99% the speed of light and so if you put 2 TVs back-back the electrons are moving apart at almost 2c

3. Feb 17, 2009

### Staff: Mentor

I'm sure you didn't mean that. They can communicate just fine.

Last edited: Feb 17, 2009
4. Feb 17, 2009

### mgb_phys

I assume he meant X,Z had a speed of c/2 + 1 relative to the origin not each other.

5. Feb 18, 2009

### marcus

Slight correction. No reason they can't communicate. Even moving in opposite directions, given enough time, light from X can catch up with Z.

With Lexus' assumptions about the speed, suppose after they have each traveled one mile in opposite directions (as seen by a stationary observer at the origin) then X sends a flash of light to Z. Then Z will receive the flash of light when each has traveled 2 additional miles.

If you think of X going west and sending the signal when he is at -1 and Z is at +1, then the signal will arrive when Z is at +3. The light goes approx twice as fast as the objects so it will have gone the 4 units from -1 to +3 in the same time it takes for Z to go from +1 to +3.

Think of it from the standpoint of the stationary observer. He sees the flash emitted and travel (in his reference frame) from -1 to 3.

The same is true even if they each travel 99% of c, in opposite directions. Given enough time, the flash of light will always catch up to

Z since Z is traveling at less than the speed of light.

Last edited: Feb 18, 2009
6. Feb 18, 2009

### mgb_phys

Doh - yes, sorry I was thinking of something else.
As soon as the light leaves X it is moving at C in the origin reference frame.

7. Feb 18, 2009

### Lexus Dominus

cheers for the help!

i think im beginning to understand, though it is quite hard to comprehend. Does light, always only move at c in accordance to a centralised reference frame (the frame of Y, the source of the whole system) i.e the reference frames of both X & Z are irrelevant, and moving objects have nothing at all to do with electromagnetic radiation (aside from space-time curvature ect). I assume this reference frame must be the origin of the big bang? is there a name for this universal reference frame of C? i gather this means the light is also hitting Z at exactly C and the wavelength of C is not affected by the speed of the object it is leaving/impacting.

Ok, so what if you reset this environment, dump a load of heavy elements in there as well & do the same thing? could the EM radiation be slowed down (by passing through substance) to the point at which it would never reach Z, assuming the stability of this environmnent? ; to rephrase, if i shone a torch through the atmosphere into space, would it speed up as it entered vaccum to C? Assuming an instant change from gas to the vaccum of space, would there be acceleration time or would it be instantaneous?

Apologies for all of the questions guys, & thanks again for your help its much appreciated.

Last edited: Feb 18, 2009
8. Feb 18, 2009

### Staff: Mentor

Hehe, no problem. I figured it must've been a late-night post.

9. Feb 18, 2009

### Dmitry67

No, you are not starting to understand :)

There are no 'special' rest frames, in SR all frames have equal rights
Speed of light is C in *any* frame.

In the rest frame of X, speed of Y is, say, 0.99, but speed of Z wont be 1.98, as you expect. If will be less then C even grater then 0.99c
Addition of velocities in relativity does not work they way you got used to.

Last edited: Feb 18, 2009
10. Feb 18, 2009

### Lexus Dominus

Ok, i am not beginning to understand, i am completely confused.

could you tell me the formula that is used to calculate the relative speeds of the objects in this scenario?

11. Feb 18, 2009

### Staff: Mentor

Here is the HyperPhysics page on http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/einvel2.html" [Broken]. As Dimitry67 mentioned, you can use this formula to determine that something which travels at c in one frame travels at c in any other frame also.

Last edited by a moderator: May 4, 2017
12. Feb 18, 2009

### Lexus Dominus

ahh thankyou.. thats what im after :)

13. Feb 22, 2009

### v2kkim

In relativistic physics, which is the reality, the coordinate frame is critical. In the example, x is moving to west and z to east fast and z is receiving a light from x, we think it may take long time for the light to reach z, because z moves very fast, but remember that here we are using the y as a reference frame. Now if we take z as a reference frame, then we can say the light will get to z very quick because the light is coming to z at full speed of light, so the time concept changes quite depending on the reference frame choice, and all relativistic. So the time elapsed from x to z can be long and also short -- both are correct, which is an interesting but critical fact of relativity.

14. Feb 22, 2009

### HallsofIvy

If one object is moving at speed u with respect to a given reference frame, and another object is moving in exactly the opposite direction with speed v, then the speed of either one, relative to the other is
$$\frac{u+ v}{1+\frac{uv}{c^2}}$$
In particular, if u= v= (1/2)c+ 1 then u+ v= c+ 2 but uv= (1/4)c2+ c+ 1 so that uv/c2= (1/4)+ 1/c+ 1/c2. Since 1/c and 1/c2 are extremely small, we can drop them without error (in the first few thousand decimal places) and have
$$\frac{c+ 2}{1+ 1/4}= \frac{c+2}{\frac{5}{4}}= \frac{4}{5}(c+ 2)< c$$.

15. Mar 10, 2009

### Karl G.

Yes- you must use Lorentz addition of velocity instead of the Galilean formula.

16. Mar 14, 2009

### Nickelodeon

To communicate in any practical sense you need wavelength and frequency. The wave length of the 'light' from X as observed from Z will be infinite, ie. redshifted to extreme. The bit that I can't figure out is that Y will be able to communicate with X, get the communique and then retransmit to Z but I'm sure this must violate GR somehow.

17. Mar 14, 2009

### Staff: Mentor

No, the redshift will be finite. In fact, the recieved frequency will be about 1/3 of the transmitted frequency.

18. Mar 14, 2009

### Nickelodeon

If the speed of separation was 2c what would be the reduction in the frequency then?

Last edited: Mar 14, 2009
19. Mar 14, 2009

### Staff: Mentor

You cannot have a speed of separation of 2c, it must be strictly less than 2c.

20. Mar 15, 2009

### Nickelodeon

Ok - what about the frequency reduction for 1.9c?

21. Mar 15, 2009

### Staff: Mentor

Assuming that the separation speed of 1.9 c is because one is going at 0.95 c and the other is going at -0.95 c then the recieved frequency is 0.026 times the emitted frequency. On the other hand if one is going at 0.99 c and the other is going at -0.91 c then the recieved frequency is 0.015 times the emitted frequency.

22. Mar 16, 2009

### Nickelodeon

Thanks for that - could you let me know how you worked it out? It seems a bit strange the difference between the two scenarios because as far as X and Z are concerned they could have no external reference frame to determine who is going faster and I thought in GR it's the relative speed that is important.

23. Mar 16, 2009

### JesseM

If they are going at 0.95c and -0.95c in the frame of the third observer Y, then you can use the velocity addition formula to work out that in the rest frame of either one of them, the other one is moving at (0.95c + 0.95c)/(1 + 0.95^2) = 1.9c/1.9025 = 0.998686c. It's the relative speed of one of them in the other one's rest frame that determines the Doppler shift that one of them sees when he gets a message from the other one (so you can see that their speed in Y's frame is in fact irrelevant here, as you said 'it's the relative speed that is important'); plug -0.998686c into the relativistic Doppler effect formula (negative since positive v in that equation represents an approaching source) and you get $$\sqrt{(1 - 0.998686)/(1 + 0.998686)}$$ = 0.02564, so the observed frequency is equal to the emitted frequency times 0.02564.

24. Mar 16, 2009

### Staff: Mentor

Sure. You are exactly correct that it is the relative speed that is important. But don't forget, the speed of separation in some reference frame is in general not equal to the relative speed. To calculate the relative speed you have to use the http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/einvel2.html#c2" formula. So for -.95 c and .95 c the relative velocity is 0.998686 c, but for -.99 c and .91 c the relative velocity is 0.999526 c.

Once you have the relative velocity then you use the http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/reldop3.html#c3" formula to obtain the frequency shift. Don't forget on the Hyperphysics page that positive velocities are approaching, so for this scenario you need to enter negative velocities.

Last edited by a moderator: Apr 24, 2017
25. Mar 16, 2009

### Nickelodeon

One last question -

If there are galaxies receding from us with a speed of separation (from us) at, say, 1.5c,
do you think it would be theoretically possible to catch them up? Sort of inferring that we can travel faster than light because they are.

Last edited: Mar 17, 2009