# Issues with relativity

1. Jun 9, 2013

### phillip1882

I've tried to understand the theory of relativity several times and could never wrap my head around it. so i would like to make an offer here. i would like to chat with a physics professor online for about an hour over Skype. my discussion probably does not belong a serious physics forum which is why i am not posting my questions here. if you would prefer that, I don't mind posting my questions here instead.
if I'm breaking any forum rules i would like to apologize ahead of time, and just say I'm not quite sure where else to turn.

Last edited by a moderator: Jun 10, 2013
2. Jun 9, 2013

### phyzguy

Why not just post your questions here? I'm sure people would be happy to try and answer them. You know what they say, the only stupid question is the one that is not asked!

Last edited by a moderator: Jun 10, 2013
3. Jun 9, 2013

### phillip1882

if i have two light beams, traveling the same direction, both going the same speed (because the speed of light is constant) what speed would one light beam be traveling relative to the other light beam? the only logical conclusion i personally can come to is: zero.
again let's say i have two light beams, this time i fire them at each other. both light beams again are traveling at the speed of light, so the speed of light of one light beam relative to the other must therefore be 2*the speed of light. if I'm wrong I don't quite see how. let's start of with this.

Last edited by a moderator: Jun 10, 2013
4. Jun 9, 2013

### Mentz114

Velocity composition in relativity uses a rule which ensures that the 'added' velocities cannot exceed c. Actually, one cannot use light to measure a velocity against - it will always be travelling at c regardless of the matter frame. If two beams are travelling parallel, one could say ( pace above ) that they had zero relative velocity.

Last edited by a moderator: Jun 10, 2013
5. Jun 9, 2013

### VantagePoint72

Yes, in your frame of reference the light beams are traveling apart at 2c. That's fine, and doesn't contradict relativity: relativity says that the speed of light is c in all inertial frames. Both beams of light are traveling at c in your frame. The subtleties of SR come in when you try to convert relative velocities between frames. Now, good inertial frames of observers in SR must travel strictly less than the speed of light, so to simplify things let me change your example slightly. A high energy particle is traveling at 0.8c to my left, and a second one travels at 0.8c to my right. In my frame, I would say the relative velocity between the two is 1.6c. Again, this is fine—both velocities relative to me are less than c.

Now, what about the frame of reference of the left-going particle ("particle 1")? How fast is the right-going particle ("particle 2") moving away according to particle 1? This is where our pre-relativistic intuition doesn't serve us very well. The answer is not 1.6c again because relative velocities add differently when transforming between frames. The formula is this, and it gives about 0.98c in this case. So, according to particle 1, particle 2 is moving less than c (as it must!). The reason for this is that velocity is the ratio of distance traveled to time, but time dilation and length contraction mean that this ratio can depend on your frame of reference. If you play around with the velocity addition formula, you'll see that if you plug in any two speeds less than the speed of light, you get a relative speed in one of the moving frames that is also less than the speed of light.

Last edited by a moderator: Jun 10, 2013
6. Jun 9, 2013

### phillip1882

okay i'll try again. let's take the example of two particles, both going 0.8*the speed of light form the position of an inertial observer.
if they are going in the same direction, what speed would one particle be going relative to the other?
again, i would say 0. if they are going opposite directions, they must be going 1.6 relative to one another.
now the previous poster has said this is wrong, that is, if one of the 0.8 particles had eyes, he would see the other 0.8 particle going 0.9756.
okay I'm very curious as to how exactly this would occur.
so let's do some math.
let's say we have a room that's 8 light minutes long. both particles start in the middle.
for a particle traveling at 0.8*the speed of light, to go from the half way point to the edge of the room would take 5 minutes.
the time dilation equation is: sqrt(1-v^2/c^2).
the space dilation equation is sqrt(1-v^2/c^2).
so, from the perspective of the 0.8 particle, he would see himself approaching the wall at:
new time: 5/sqrt(1-0.8^2) = 8.33 minutes.
new distance: 4/sqrt(1-0.8^2) = 6.67 light minute length.
final velocity = 0.8*the speed of light.
so he doesn't see any change in his own velocity relative to the stationary observer; even factoring in space and time dilation. so, can you show me how he would see the particle moving away from him as anything other than 1.6*the speed of light?

Last edited by a moderator: Jun 10, 2013
7. Jun 9, 2013

### Staff: Mentor

No, he sees the wall approaching him at .8c while he's at rest.

The other particle also sees itself at rest, while the wall is moving away at .8c.

And the particles both see themselves at rest while the other particle is moving away at a speed that is not .8c+.8c, but rather is $\frac{(.8+.8)c}{1+(.8)^2}$
The left-moving particle sees the left-hand wall approaching at .8c, but the right-moving particle does not see the distance between the wall and the left-moving particle (both moving away from it) closing at that rate.

8. Jun 9, 2013

### ghwellsjr

You got the right numbers for the time and distance the wall has traveled in the particle's rest frame but instead of calculating the factors in isolation and trying to interpret them just use the Lorentz Transformation on the coordinates of the rest frame of the walls and draw a diagram of the rest frame of one of the particles. Can you do that?

To make it a little easier for you, I'm providing a diagram for the rest frame of the two walls (shown in black). The two particles are shown in red and blue. The dots mark off one-minute increments of Proper Time for each object:

Now to use the Lorentz Transformation with a speed of 0.8c to get the coordinates for the rest frame of the blue particle. You only have to transform the coordinates of the five events at the extremities of each worldline. Then mark off the events on this template and draw in the worldlines (don't worrry about the dots):

#### Attached Files:

File size:
2.1 KB
Views:
271
• ###### TwoParticles@.8C.PNG
File size:
1.9 KB
Views:
275
Last edited: Jun 10, 2013
9. Jun 10, 2013

### Fredrik

Staff Emeritus
How much math do you know, philip1882? Are you familiar with matrices and matrix multiplication for example?

I suggest that you start by familiarizing yourself with spacetime diagrams, as explained by Schutz here. Unfortunately not all the pages of the explanation are viewable online. But maybe you can get the book from a library or something.

10. Jun 10, 2013

### phillip1882

http://postimg.org/image/d5ug05njf/
wouldn't this be it?
i guess i don't quite understand why the equation for adding velocity is what it is, given that velocity doesn't change relative to the stationary observer.

11. Jun 10, 2013

### Staff: Mentor

It follows directly from the Lorentz transforms, which in turn follow directly from Einstein's two postulates. You'll find derivations all over the internet, for example, here: http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/veltran.html

12. Jun 10, 2013

### D H

Staff Emeritus
The simple answer is that it doesn't. Multiple experiments demonstrate that the simple vector addition of velocities just doesn't apply to objects moving at very high velocities.

One key thing that physicists have learned over the last 100+ years is that the nice and simple world of Newtonian mechanics does not extrapolate to the world of the very small, very large, or very fast. Extrapolation is always at best a dubious thing to do. It turns out to be a very bad thing to do in physics.

In our mundane, ordinary world, velocities apparently do add vectorially and passage of time apparently is the same to all observers. These apparently obvious facts are observably false in the world of the very, very fast. The failures of these cherished concept is itself a fact. Your choices are to
• Accept these new facts as facts and learn from them, in which case we'll be quite happy to try to help you along that road, or
• Reject these facts as apparent nonsense, in which case we'll be quite happy to tell you to peddle your nonsense elsewhere.
The choice is up to you.

13. Jun 10, 2013

### atyy

In Newtonian physics, we usually use inertial coordinates, which are the "common sense" coodinates with spatial axes at right angles and time such that Newton's second law holds. But although they may seem common sense, they are actually coordinates that respect an underlying symmetry of Newtonian physics.

In special relativity, the concept of inertial coordinates also exists, but the symmetry is slightly different. If the "stationary" observer sets up an inertial coordinate system, then the "moving" observer is also able to set up an inertial coordinate system - both of them use the same procedures to set them up. However, the relationship between the "stationary" and "moving" inertial coordinate systems is different from Newtonian physics, because of the different underlying symmetry of special relativity. This is why the velocity addition formula is different in special relativity.

In a sense, setting up inertial coordinates in special relativity defines the speed of light to be the same constant in all inertial frames. So it is a definition, but it ultimately makes things simpler because it respects the underlying symmetry of the theory - which also happens to be a much better description of the symmetry that is present in nature than the symmetry of Newtonian inertial coordinates.

14. Jun 10, 2013

### phyzguy

Phillip1882,

Let's try and complete the exercise GHWellsJr started and maybe this will help you understand. In the frame where the two walls are at rest, and the two particles are moving left and right at 0.8c, the space-time diagram looks like the attached (Rest_Frame.png). The two walls are in black, and the two particles are in red and blue. The LH particle is watching the RH particle, and when the RH particle reaches the wall (at X = 4.0 lightminutes, T = 5.0 minutes in this frame), a light ray (green line) travels back to the LH particle, arriving at the LH particle at X = -36 light-minutes, T = 45 minutes.

Now let's look at it from the frame of the LH particle, which is shown in the attached LH_Particle_Frame.png. Because of Lorentz contraction and time dilation, the LH particle sees the two walls (in black) each 2.4 light-minutes from the origin (the rest-frame 4.0 light-minutes times the Lorentz contraction factor of 0.6), and each moving to the right at 0.8c. The light ray emitted from the RH particle when it reaches the RH wall arrives at the LH particle at T' = 27 minutes (this is T=45 minutes in the wall rest frame times the time dilation factor of 0.6). Extrapolating the light ray backwards along the green line, the LH particle concludes that the RH particle arrived at the wall at T' = 13.67 minutes, and that the RH particle traveled 13.33 light minutes in this time (because it had to travel the original 2.4 light-minutes + 13.67 minutes * 0.8c to reach the RH wall). So, he concludes that the RH particle traveled 13.33 lightminutes in 13.67 minutes, giving a velocity of 0.975c, which is just what is given by the velocity addition formula of (0.8+0.8)/(1+0.8*0.8) = 0.975.

Does this help?

#### Attached Files:

File size:
8.4 KB
Views:
72
• ###### LH_Particle_Frame.png
File size:
10.1 KB
Views:
72
Last edited: Jun 10, 2013
15. Jun 10, 2013

### Naty1

yeah, well join the club....

Before an 'Einstein', neither did the finest physicists in the world understand! "experts' in the early 1900's were trying to understand Maxwell's equations and were making some progress via Lorentz and Fitzgerald, but only Einstein could put the theory together.

atyy posts:
In retrospect, even Einstein did not fully understand that spacetime is a four dimensional continum....his college math professor [Herman Minkowski] had THAT insight....so Einstein put the algebra together for special relativity based in part on the work of Maxwell, Lorentz and Fitzgerald...but Minkowski provided the geometric view of a four dimensional spacetime.....Einstein adopted that view and used it to develop general relativty....curved spacetime!! Utterly extraordinary....

Minkowski's famous comment:
In other words, with a fixed speed of light, not some ficticious 'ether' which was suspected in that era, space and time conspire jointly, changing in flat spacetime between the two so as to maintain the constant speed of light. When time dilates, space contracts...and 'c' stays constant.

Nobody quite understands THAT either...it's not like Einstein started from some fundamental first principles...he took available mathematics and experimental observations...and somehow found out what matched. All we know for sure is that, so far, such a relationship matches observations.

16. Jun 10, 2013

### phillip1882

hmm. This does help me understand how you're deriving the equation, but i must confess i have a problem with this derivation. you're using light itself to determine the speed of two objects that could potentially be moving faster than the speed of light away from each other.
i mean, imagine i did the following. let's say i used two baseballs moving apart at the rate of 100 miles an hour (combined, so 50 miles apiece), and then i tried to measure the speed between them by using a 60 mile an hour ball. you understand i may get a vastly different velocity than whats actually occurring?
so, let me try asking the question i have again. if, after factoring in time and space dilation, the velocity of an object remains the same, how can the velocity of two objects not simply add and subtract?

17. Jun 11, 2013

### Staff: Mentor

I'm not sure what exactly what you mean by "the velocity of an object remains the same" (remains the same as what?), but if you actually calculate the effects of time dilation and length contraction you get the relativistic formula. I pointed you to one such derivation in post #11, but could probably find another derivation out there that uses no calculus if you would prefer.

But sometimes... There really is no substitute for doing the math.

18. Jun 11, 2013

### jartsa

The correct calculation of final velocity:

new distance: 4*sqrt(1-0.8^2) = 2.4 light minute length.

new time: 5*sqrt(1-0.8^2) = 3.0 minutes.

final velocity = new distance / new time = 2.4 / 3.0 = 0.8 *speed of light

Last edited: Jun 11, 2013
19. Jun 11, 2013

### Staff: Mentor

You should work out the math. You would get the right answer. In fact, you would still get the right answer if the measurement balls were going 10 mph.

The one difference is that the speed of baseballs is not frame invariant, so you could only do this procedure in the frame where your measurement balls speed was known.

Last edited: Jun 11, 2013
20. Jun 11, 2013

### phyzguy

It's not a derivation. It's a description of how the universe behaves, first worked out by Einstein, and verified by a huge body of experimental data until it is currently beyond doubt. The description assumes as one of the postulates that the speed of light is the same in any reference frame. This is why light plays such a central role.

For baseballs moving at 50 miles an hour, the difference from simple velocity addition is extraordinarily tiny. For two baseballs moving apart at 50 miles per hour, an observer sitting on one baseball would measure the other baseball as moving at 99.9999999999994 miles per hour, an immeasurably small difference from 100 miles per hour.

It's certainly not the same. In the rest frame of the walls, the LH particle is moving at 0.8c and the walls are stationary. In the rest frame of the LH particle, the velocity of the LH particle is zero and the walls are moving at 0.8c. The number is the same, but the physical situation is very different.