How Does Relativistic Motion Affect Batman's Batship Shadow and Laser Image?

[yuiop]

Imagine Batman has a new hyper speed Batship that is the shape of his circular bat emblem. The ground has convenient square grid on it and when parked on the ground the ship is exactly the same length as a square on the grid and its shadow touches the four sides of a square. Mounted underneath the Batship is a series of powerful lasers pointing directly downward and when fired simultaneously they burn an image of the Batship on the ground that has the same profile as the shadow.

Now Batman takes off and passes low over the ground at a relativistic speed such that gamma = 2. It is high noon and the shadow projected on the ground is half the length of the squares on the ground grid. He fires his lasers briefly and simultaneously and burns an image of the craft on the ground. The burned image on the ground is twice the length of a square grid and 4 times the length of the cast shadow seen by an observer on the ground.

So, do you agree with this conclusion and do you see the "trick" that brings about this seemingly contradictory conclusion?

If that is too easy, try this. What is the length of the shadow and the burned image relative to the squares on the ground according to Batman as he flies past?

 

[Goodison_Lad]

I think I agree. When he fires his lasers 'simultaneously', they don't fire simultaneously in the ground frame - the rear one fires first. By the time the front one fires (in the ground frame) the ship has traveled further than its rest length.

For the second question, if the ground measures the shadow to be half of the grid length then, by the symmetry of the relative velocity, Batman must measure the length of the grid to be Lorentz contracted to half of his ship's length. And since the burned image is twice the size of the grid, he'd measure the burned image to be the same length as his ship.

As for the shadow, the ground would determine its length by measuring its front and rear locations at the same intant in the ground frame. These instants would be asynchronsied in Batman's frame. Not sure yet what he 'see'. I'll mull this over.

 

[yuiop]

Hi GL, I think you have discerned the "trick" correctly as being due to the relativity of simultaneity.

As for the shadow, the ground would determine its length by measuring its front and rear locations at the same intant in the ground frame. These instants would be asynchronsied in Batman's frame. Not sure yet what he 'see'. I'll mull this over.

My guess is that Batman measures the shadow and the burned image to both be the same length (and equal to the length of his ship) and twice the length of a square on the grid.

So according to an observer on the ground, the shadow is shorter than the burned image and to Batman on the moving ship the shadow and the burned image are the same length. Both observers agree the burned image is twice the length of a square on the grid but disagree about the length of the shadow.

 

[Goodison_Lad]

This is a good question - somebody should put it in a book.

Yes, the more I think about it the more I agree he'd see the shadow to be the same length as his ship. It would be strange indeed if Batman saw his shadow change in size - this would give him a way to tell that he was 'really' moving, and that would be rude.

I'll have a go at dusting off my calculation brain to double-check.

 

[mfb]

Well, as we have light from some direction, Batman could measure his own speed relative to the ground in terms of the angle (and frequency) of the incoming light.
But you are right: For Batman, burning an emblem and looking at the own shadow are identical things - just imagine that his lasers "replace" the light blocked by the ship for a moment (in bat frame).

It is a nice feature that different observers can disagree about the length of a shadow, but if you consider that the shadow is moving it can be explained.

 

[PAllen]

Well, as we have light from some direction, Batman could measure his own speed relative to the ground in terms of the angle (and frequency) of the incoming light.
But you are right: For Batman, burning an emblem and looking at the own shadow are identical things - just imagine that his lasers "replace" the light blocked by the ship for a moment (in bat frame).

It is a nice feature that different observers can disagree about the length of a shadow, but if you consider that the shadow is moving it can be explained.

In further agreement with above, a shadow is an immaterial (not even energy) pattern. It is analogous to an inverse shadow - the spot of light from a flashlight. It can move faster than light, and if you look at the its process of formation (e.g. a light is turned on, soon producing a shadaw) it will be frame dependent which end of the shadow forms first.

 

[yuiop]

Is this the first thread where no one has disagreed with anyone else by post #6?

 

[Adel Makram]

The burned image on the ground is twice the length of a square grid

so I see the solution has 3 steps,

1) the grid size is 1/2 the ship size as measured by batman
2) the burned image is twice the grid size as measured by the ground observer
3) so, the ship size is equal to the burned image as measured by batman

I just don't understand step 2!

 

[omega_minus]

I agree with the above answers as well. This puzzle reminds me of the one with the snake moving close to c across a table and two hatchets (one snake width apart) falling simultaneously in the tables frame as he passes under them. Since the snake is contracted in the table's frame while moving he avoids being chopped, but in the snakes frame the hatchets are closer together than he is long! Of course he avoids being cut in both frames, its just that (like the front and back of Batman's shadow) the hatchets fall at different times to the snake.

 

[Adel Makram]

2) the burned image is twice the grid size as measured by the ground observer


I just don't understand step 2!

Ok I got u,,,
First determine the length of burned image on the ground. Because laser is fired simultaneousely, so t'2=t'1
X2= γ(x'2 +v t'2). and x1= γ(x'1 +v t'1)
X2-x1= γ(x'2-x'1) = γ *(the rest length of the ship ).,,,,,,, (1)
Now the burned image on the ground is seen by the batman at t1'=t'2 leading to a Lorentz contraction formula ,,, L'= 1/γ ( x2- x1),,,,,,,(2)
From 1 & 2 ,,, L'=1/γ * γ (x'2 - x'1) = x'2 - x'1
Which mean the length of the burned image on the ground as measured by batman is equal to ( L') the length of the ship