# Lorentz, simultan, radar finds dust grain, inconsistencies

1. Apr 17, 2013

### Albertgauss

Hi all,

Here is the problem. I want to try to do some calculations with seeing how much time a relativistic ship has before it can dodge a dust grain. I am having some trouble applying the Lorentz Contraction Equations to get consistent results.

Take two frames, the coordinates of the dust grain at rest with respect to the Earth, and a ship moving with v near c. In the below, exponents marked with a "^". In all coordinates, position is first, time is second.

At x=0, t=0, x'=0, t'=0 the Earth and ship origins line up.

For the Earth frame, the dust grain is at (d, 0).
If you do the Lorenz Transformation Equations, you will find that in the ship's frame, the
coordinates transform to: (d', -γvd/c^2) where the last term is the time offset in the ship's frame. d' = γd. This part is correct, and is referenced in Griffiths 3rd Edition.

Now, the Earth says that at its x=0, t=0, the ship sends out a radar pulse. This radar pulse travels at c, being a photon, and it will arrive at the dust grain at (d, d/c).

If I do a Lorentz Transform at the event on when the dust grain recieves the radar pulse, I get the following:

equation 1 equation 2

d' = γ(d-vd/c) and t' = γ(d/c-vd/c^2).

γ is gamma, v is velocity, d is dust grain coord for Earth, c is light speed. d' is grain coord as seen by ship, t' is time coord as seen by ship for grain to recieve radar pulse.

Notice that in the ship's t' calculated here, the -γ(vd/c^2) term matches the transformed t' from transforming (d,0) to the ship's time.

Now, here is the problem. Look closely at t' = γ(d/c-vd/c^2). It has two terms. The way I read this is that, in the ship's frame, at the time beginning with the offset -γ(vd/c^2) the ship fired a radar pulse at c and the radar pulse travelled a distance d'/c. The problem I have here is: doesn't the dust grain move towards the ship, so that if I didn't use a Lorentz Transform from a ground frame, I would write d'-vt'=ct' (equation 3) for the ship? If, I discard th " vt' " term in equation 3, I agree with the first term in equation 2 above. In writing d'-vt'=ct', I think I am saying that my radar pulse will collide with the dust grain moving towards me, but this doesn't agree with equation 2. Why doesn't this make sense? The first term of Equation 2 seems to say that the dust grain won't move at all in the ship's frame, but simply be contracted. Equation 2 seems to imply that in the ship's frame, the radar pulse will travel to the dust grain in time d'/c. So I don't know which to believe here.

I looked up some derivations of the lorentz transform. Is the idea that, in the ship's frame, the ship stays still and the ground moves with respect to the ship already incorporated into the structure of the the Lorentz Transforms? This is all I could think of to reconciles this inconsistency.

Also, when the radar pulse reaches the dust grain, I tried to figure out where the Earth and ship say the new coordinates of the ship are. For the Earth, this is easy. In the Earth's frame, the ship will be at (vt, t). The ship will move to a new coordinate vt, where t is the time it sees the ship's radar pulse reach the dust grain, i.e, t=d/c.

If I transform to the ship's frame, I get x_s'=0 which makes sense; the ship percieves itself to be at its own origin even though Earth says the ship is at x_s = vt. This is mostly a check to make sure I am doing things right.

But if I transform the time from Earth to ship, shouldn't I get the same transformed time of equation 2? The ship is at x_s = vt at time "t" and that is the same "t" the dust grain recieves the radar pulse in the Earth's frame. However, because the x coordinates of the ship and dust grain are different, I don't. Here's what happens:

t_s'=γ(d/c-v*vt/c^2)=γ(d/c-d*(v^2)/(c^3)).

The first term in this transformation is right, but the 2nd term -d*(v^2)/(c^3) does not at all match the second term in equation 2. If the ship, stays at his origin and watches the dust grain absorb the radar pulse, shouldn't he have the same transformed time for both events? Shouldn't the Earth transform to the same time in the ship's frame for both events? Or have I got caught by simultaneoity here? The ship may be at x_s = vt and time "t" and the dust grain may also recieve the radar pulse at the same time "t" in the Earth's frame, but if I go to the ship's frame, will a simultaneity issue not let the transformed times be the same? That seems very weird.

2. Apr 19, 2013

### ghwellsjr

Rather than respond to each of your points, I'm just going to make some general statements and then illustrate your scenario with a specific example and a couple of spacetime diagrams.

1) I don't know why you want to parse the terms in the Lorentz Transformation. It works just fine the way it is.

2) I don't know why you want to use the Lorentz Transformation process to determine how much time a spaceship has to avoid a dust grain that it is on a collision coarse with the spaceship. Any frame will do and especially the one in which you describe the problem.

3) You seem to think that different frames provide different information to the observers that are stationary in them but this is a common misconception.

4) It is very difficult to appreciate what the Lorentz Transformation is doing to the coordinates of events unless you plot them and that is because when you are looking just at numbers, it is not obvious how time is progressing. When you plot the events on a spacetime diagram, their organization becomes obvious because you no longer see them in terms of the original defining frame but in terms of the new frame.

Now for the example. I'm going to use a speed of 0.8c for the spaceship and we'll put that dust grain at one light-year away from earth and stationary with respect to the earth. The spaceship is on a collision course with the dust grain and it sends a radar pulse to the dust grain as soon as it leaves earth.

Here is a spacetime diagram depicting the earth in blue and the dust grain in red:

Both are stationary in their common rest frame. The spaceship in black is traveling at 0.8c from the earth toward the dust grain. The radar pulse is depicted as a green line traveling along a 45-degree angle. As you can see, it takes one year (of Coordinate Time in this frame) for the pulse, traveling at the speed of light to reach the dust grain and a month (of Coordinate Time in this frame) for the reflection to get back to the spaceship. However, because the spaceship is traveling in this frame at 0.8c, gamma for it is 1.667 which is the ratio of Coordinate Time to its own Proper Time. In Coordinate Time, it takes 15 months for the spaceship to travel 12 light-months (12/15=0.8) but the spaceship's Proper time is 15/1.667=9 months. The dots for each observer/object mark off one-month intervals of Proper Time. The spaceship receives the radar echo at month 8 (of its Proper Time) and it has one more month (of its Proper Time) to avoid colliding with the dust grain.

Now that is all you need to know to answer your question. However since you want to use the Lorentz Transformation process to get to the rest frame of the spaceship, I have provided that spacetime diagram:

I hope you can see that even though the coordinates are completely different, the observations and measurements of each observer/object according to their own Proper Times are identical to what they were in the first diagram. Can you see that even though the radar takes only four months to get to the dust grain and another four months to get back to the spaceship, the spaceship still receives the echo at the same time that we already determined, namely after eight months (of its Proper Time), leaving one month (of its Proper Time) to avoid collision?

Hope this helps even if I didn't address your specifics but I think you will be able bridge the gap and fill in the details for yourself.

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3. Apr 19, 2013

### Staff: Mentor

For convenience I will use units where c=1.
That is actually just a single event on the dust grain's worldline. The dust grain exists over time, so a better expression would be that the dust grain is at (x,t) = (d,τ). This can be seen as a parametric equation of a line, also known as the worldline, where τ is the parameter along the worldline.

Using the formulation above you get (x',t') = (γ(d-vτ),γ(τ-vd))

So, if we examine (x,t) = (d,τ) = (d,d) then we immediately find that the radar pulses arrival is the event on the dust worldline where τ=d.

(x',t') = (γ(d-vτ),γ(τ-vd))
Then substitute in τ=d to obtain
(x',t') = (γ(d-vτ),γ(τ-vd)) = (γ(d-vd),γ(d-vd))
which is the same as what you found above.

Yes, it does.

The correct expression is: (x',t') = (γ(d-vτ),γ(τ-vd)). That gives you two equations so you can eliminate τ if you want. If you do that then you get:
d' - vt' = x' + v²d'
where d' = γd.

I don't know how you got your expression, but it seems that if you use the correct expression then everything else should be resolved.

4. Apr 19, 2013

### Albertgauss

Hey all,

I am reading and thinking through all this. I am in-tune with what everyone is saying. It might take me a few days to come up with a better response, but I just wanted to mention that a lack of response is not me being lazy or flippant. I am really interested in getting all the specfics and details right.

5. Apr 29, 2013

### Albertgauss

I understand now

I think I understand everything. Let me finish up with a few points.

) I don't know why you want to parse the terms in the Lorentz Transformation. It works just fine the way it is.

I was parsing the lorentz terms to understand what they mean. For instance, in the time equation, for the simultaneous event of (0,0) and (d,0) which starts everything rolling in the Earth's frame, the Lorentz transform in time is a combination of time dilation where both frames would be at the origin at the same time, but the transformed event (d,0) yields an offset in time. Yes, I understand the Lorentz equations are fine as they are; it was an issue to understand them.

The spacetime diagram really helped. How hard, or where did you find the code to write them and make that plot? Is is just a simple computer programming exercise?

2) I don't know why you want to use the Lorentz Transformation process to determine how much time a spaceship has to avoid a dust grain that it is on a collision coarse with the spaceship. Any frame will do and especially the one in which you describe the problem.

For relativity, it seems easiest for me to start with the Earth frame. I probably should work on that, but its just easiest to think of it that way.

Also, I do understand the info is the same for everywhere. If an event happens in one frame, then it happens in the other frames. Its the coordinate and time in which the event happens that frames disagree on.

I understand my error in equation 3: d'-vt'=ct'. The error is that this is what one would write for the Earth frame where the ship travels to the right, and the reflected pulse travels to the left and then the ship meets the pulse. I mislabeled the variables d' and t'. I realize now, in the ship's frame, the dust grain heads at the ship at vt' and sends the radar pulse a distance ct' before it hits the ship. In that case ct'=x'+vt'=x'+v^2d'. So I sorted this out now. (x' would then be the new coordinate of the dust grain in the ship's frame).

Yes, I have worked with that plot and it has been insightful. I got all the other details.

Thanks for your help all, I think I am good to go.

6. Apr 30, 2013

### ghwellsjr

Did parsing the equations help you to understand them?

I'm glad they helped you, they help me, too.

I wrote the application in a language called LabVIEW which I use professionally, during the course of responding on the thread called Triplet Paradox. When all the features are enabled, it draws a diagram like this (from post #140 of the aforementioned thread):

The hardest part of programming the application was getting the two axes to have exactly the same scale so that light would be depicted as traveling on 45-degree diagonals. I can change the size of the diagram which was part of the challenge.

I can disable different features, such as the light signals or even one or both of the outside triplets. This allows me to use the application for Twin Paradox or other simple scenarios. You will note that the light signals in this diagram start at the Proper Time "ticks" for the three triplets and continue forever (actually, just a long distance) whereas in the diagrams I made for you, the light reflected light signal just goes between the objects. That's because I disabled the light signals in the application, copied the image and edited it in Paint. Making the light signals go forever was easy. If I had made them stop at an object or reflect, that would be a big challenge.

Actually, I had to do a lot more editing than that because your scenario was way different, so essentially, I just used my application to create portions of your diagrams and put it all together in Paint. One of these days, I plan to write a more general application that will allow me to specify any scenario and then use the Lorentz Transformation process to generate the other diagrams.

I agree and I also used the Earth frame to begin with too. Nothing wrong with that. If you have an automatic process to do the Lorentz Transformation, then it's enlightening to do it to see that it has no effect on what the different observers/objects/clocks can see, measure or observe.

You're very welcome. We like success stories and appreciate the positive feedback, which believe it or not, is somewhat rare.