Finding Angle Theta in Relativistic Skydiving Scenario

[jlmccart03]

Homework Statement

A skydiver is strapped to a rocket and shot into outer space. There is no air resistance so he is able to make his arms into a 45° angle with respect to his body and pretend they are "wings." To a stationary observer on the Moon (we will consider the Moon stationary for our purposes), at what angle will it appear that he is holding out his arms with respect to his body when he goes speeding by head-first in the -x direction at speed v = 0.6c?

Homework Equations

Length contraction or Lorentz transformation

The Attempt at a Solution

I think this is a Lorentz transformation, but I do not know how to get an angle theta. I only know of time dilation and length contraction niether have a theta in their equations so what am I missing?

 

[Orodruin]

so what am I missing?

Trigonometry. Consider the arms the hypothenuse of a right triangle with the body as one side.

 

[FactChecker]

Think of a right triangle where his arm is the hypotenuse, the height is in the direction of travel and the base in perpendicular to the direction of travel. The Lorentz transformation changes the height but not the base. So the angle changes in a way that you can calculate.

 

[jlmccart03]

Think of a right triangle where his arm is the hypotenuse, the height is in the direction of travel and the base in perpendicular to the direction of travel. The Lorentz transformation changes the height but not the base. So the angle changes in a way that you can calculate.

Ok, so I have a triangle where the hypotenuse is his arm and his body is one side of the triangle. Since in his frame of reference it is a 45, 45, 90 degree triangle I am now confused on how this changes in the reference frame of the observer? So from what you said, the hypotenuse is in the -x direction since that is the direction of travel. So using the pythagorean theorum I have √(v*t)²+h² = c*t and I know v = 0.6c. Is that a step in the right direction? I have a feeling this is going to end up as two right triangles meshed together or is that wrong?

 

[Orodruin]

So from what you said, the hypotenuse is in the -x direction since that is the direction of travel

No, this is not correct. The arm is the hypothenuse and makes a 45 degree angle to the travel direction.

I have √(v*t)2+h2 = c*t

What is t doing in this equation? The question a priori has nothing to do with any particular time.

 

[jlmccart03]

No, this is not correct. The arm is the hypothenuse and makes a 45 degree angle to the travel direction.What is t doing in this equation? The question a priori has nothing to do with any particular time.

I was following a lecture slide and thought time was somehow related, but its not. So I need to find a way of having a triangle with the length contraction form of x = x₀/ϒ. Is that what I am trying to do? And I drew a triangle with the 90 angle at his head, and then an leg on his arm, a leg from his head to shoulder, and a leg from head to hand correct?

 

[jlmccart03]

No, this is not correct. The arm is the hypothenuse and makes a 45 degree angle to the travel direction.What is t doing in this equation? The question a priori has nothing to do with any particular time.

Ok, I think I solved it. I know that x = x₀/gamma. Gamma is found by taking 1/(√(1-(0.6c)²/c²) or 1.25 when solved. Then looking at the triangle I want the angle between his arm and shoulder basically. That means I take tan(θ) = opp/adj or x₀/(x₀/1.25) which is arctan(1.25) = 51.34°. Correct?

 

[FactChecker]

That doesn't sound right. I suspect that you are trying to do this in your head without drawing it out and am getting things confused. Make a drawing of the still right triangle with a 45° angle. Change the side (sides?) that should change when it is moving with respect to the Moon (draw the direction of motion). Then do some trig.

 

[jlmccart03]

That doesn't sound right. I suspect that you are trying to do this in your head without drawing it out and am getting things confused. Make a drawing of the still right triangle with a 45° angle. Change the side (sides?) that should change when it is moving with respect to the Moon (draw the direction of motion). Then do some trig.

This is my picture.

 

[FactChecker]

And which direction is he going in? It looks like your equation is for him going sideways. Also, double check what you are considering opposite and adjacent.

 

[jlmccart03]

And which direction is he going in? It looks like your equation is for him going sideways. Also, double check what you are considering opposite and adjacent.

So I changed it up and realized that tan(θ) should be x₀/-x, but I still get 51.34° just negative. Here is my picture now.

EDIT: I noticed that h=x-x₀ so I ended up getting 71° by taking arctan(x₀/x-x₀).

 

[FactChecker]

Why are both your opposite and adjacent sides of the triangle based on x₀? Don't get the sides confused. You should take a second look at the triangles and trig.

 

[jlmccart03]

Why are both your opposite and adjacent sides of the triangle based on x₀? Don't get the sides confused. You should take a second look at the triangles and trig.

Are they supposed to be based on X? I simply solved for h and got x-x₀. It simplified very nicely, but i don't know how else to set it up. I know his arm has to be x₀ since it's the proper length making the perpendicular leg x and the parallel leg h. Right?

 

[FactChecker]

I can't follow what you did. I don't see h in any equation or drawing (do you mean H in the drawing?). I can't figure out why x-x₀, which is hypotenuse - oneSide, should appear in any equation. And why is tan(θ) = oneSide/hypotenuse? That is sin(θ). I give up.

 

[jlmccart03]

I can't follow what you did. I don't see h in any equation or drawing (do you mean H in the drawing?). I can't figure out why x-x₀, which is hypotenuse - oneSide, should appear in any equation. And why is tan(θ) = oneSide/hypotenuse? That is sin(θ). I give up.

As I thought I said, I took H to be the parallel leg to his motion. I then solved for it in terms of x and x₀. The hypotenuse is x₀ the PROPER LENGTH of my Lorentz factor. That's all I did. I used trig to find h in terms of x and x₀. I have no idea what you are trying to convey. But thanks for giving up! Really helpful...

EDIT: the equation WITH h is simply tan(theta)=x/h. Now what in the world is h? I was never given anything for h.

 

[FactChecker]

As I thought I said, I took H to be the parallel leg to his motion. I then solved for it in terms of x and x₀. The hypotenuse is x₀ the PROPER LENGTH of my Lorentz factor. That's all I did. I used trig to find h in terms of x and x₀. I have no idea what you are trying to convey. But thanks for giving up! Really helpful...

EDIT: the equation WITH h is simply tan(theta)=x/h. Now what in the world is h? I was never given anything for h.

Sorry. What is 'h' in the diagram? Your calculations may be right, but I just can't follow what you are doing. Maybe someone else can understand better.

 

[jlmccart03]

Sorry. What is 'h' in the diagram? Your calculations may be right, but I just can't follow what you are doing. Maybe someone else can understand better.

No worries, this is my diagram. "h" is the leg parallel to the -x direction. It's basically the part from his shoulder to his lower head. So the neck I guess.

 

[Hendrik Boom]

Looks right to me. The foreshortening is along the length of his body, so his arms Will appear more horizontal, giving a greater angle with the body.

Except of course you want x0/H for the tangent. But except for this typo, the rest appears correct.

 

[jlmccart03]

Looks right to me. The foreshortening is along the length of his body, so his arms Will appear more horizontal, giving a greater angle with the body.

Except of course you want x0/H for the tangent. But except for this typo, the rest appears correct.

Yeah I checked with a classmate of mine and they got 51 degrees and then the solutions also showed so by my professor. I will be more clear next time as I can see many areas causing confusion. Thanks for the help everyone!