Energy Triangle: Exploring Physical Significance

[Carolina Joe]

Energy "Triangle"

It seems to me that the total energy of a particle forms a kind of right triangle.

$$E^2 = (m_0 c^2)^2 + (pc)^2$$

Where the total energy is the hypotenuse, and it's square is the sum of the other 2 sides, each squared. Does anyone know of any physical significance to this fact, or is it just a coincidental relationship that has no further meaning? I just love stuff like this.

 

[selfAdjoint]

It seems to me that the total energy of a particle forms a kind of right triangle.
$$E^2 = (m_0 c^2)^2 + (pc)^2$$
Where the total energy is the hypotenuse, and it's square is the sum of the other 2 sides, each squared. Does anyone know of any physical significance to this fact, or is it just a coincidental relationship that has no further meaning? I just love stuff like this.

Modify the equation to

$$(m_0 c^2)^2 = E^2 - (pc)^2$$

This is the magnitude of the momentum-energy four-vector expressed - by the pseudo-Pythagorean rule of Minkowski spacetime - in terms of its timelike, E, and its spacelike, pc, components. The magnitude of a four-vector is the same in all intertial frames, and the magnitude of the momentum energy four-vector is mc^2 (m being the invariant mass).

 

[Carolina Joe]

Ah, so the triangle-like quality of that expression is useful, but in an area of physics I've never studied. My texts are undergraduate level, so I guess if you're looking at graduate textbooks they go further into special realativity using Minkowski spacetime? Is that just a framework for working in relativity? I've got a lot to learn. What makes E the timelike element, and pc spacelike?

 

[ahdaf]

It seems to me that the total energy of a particle forms a kind of right triangle.

$$E^2 = (m_0 c^2)^2 + (pc)^2$$

Where the total energy is the hypotenuse, and it's square is the sum of the other 2 sides, each squared. Does anyone know of any physical significance to this fact, or is it just a coincidental relationship that has no further meaning? I just love stuff like this.

I like your question very much and it is not any kind of question. As you just guess it this so called momentum-energy relation has really no physical significance and this relationship is just coincidental..The energy E cannot be the hypotenuse but actually I've just come to the oint that p& E are perpendicular

 

[Bob S]

In the 1950's and 1960's, before the advent of electronic (and pocket) calculators, physicists used trigonometric (sin, tan, cotan, cos) tables (for 90-degree right triangles) in math books to calculate energy and momentum of particles in beamlines; e.g., E sin(theta) = pc, where beta = sin(theta). Also, cos(theta) = sqrt(1-beta²), and E cos(theta) = rest mass. Luis Alvarez I think was one of the earliest to have a HP-35 pocket calculator in the late 1960's, the earliest pocket calculator to have trig, log, and exponential functions.

 

[Jonathan Scott]

The corresponding relationship between time, space and proper time does have a geometrical triangle interpretation, as follows:

Take two space-time events and connect them in space with a piece of string of length ct, representing the time between them. The proper time between those events is then given by the slack in the string, specifically by the maximum perpendicular distance that one can move the mid-point of the string from one side to the other of the path (on the assumption of constant velocity between the events). Note that if the separation is null, the string is straight and there is no slack. If the two events are at the same point in space, the slack is equal to the length of the string and the proper time is equal to the time.

Note that if you subdivide or join these pieces of string to cover additional events, the lengths (corresponding to time) add linearly and so do the displacements between their ends (corresponding to space), but the way the slack adds depends on whether the events fall on a line of constant velocity. If they do, the sum of the slack (proper time) for each part still adds up to the total for the end events, but if there is a change in velocity involved, the total slack is less than that between the end events for the overall path.

 

[Bob_for_short]

It seems to me that the total energy of a particle forms a kind of right triangle.

$$E^2 = (m_0 c^2)^2 + (pc)^2$$

Where the total energy is the hypotenuse, and it's square is the sum of the other 2 sides, each squared. Does anyone know of any physical significance to this fact, or is it just a coincidental relationship that has no further meaning? I just love stuff like this.

In the plain (p,E) this equation describes a known hyperbola E(p).
At the same time E(p) is numerically the distance from the reference frame point (0,mc^2) to the point p in the axis p.

Bob.

 

[ahdaf]

In the 1950's and 1960's, before the advent of electronic (and pocket) calculators, physicists used trigonometric (sin, tan, cotan, cos) tables (for 90-degree right triangles) in math books to calculate energy and momentum of particles in beamlines; e.g., E sin(theta) = pc, where beta = sin(theta). Also, cos(theta) = sqrt(1-beta²), and E cos(theta) = rest mass. Luis Alvarez I think was one of the earliest to have a HP-35 pocket calculator in the late 1960's, the earliest pocket calculator to have trig, log, and exponential functions.

Dear Bob,
The trigonometric equation you give don't define the ratio E/pc of a uniquely manner.It all depends on which corner of the triangle you take.For example you can also have cos(theta)=E/pc, that means you always have to draw a triangle in order to show to the reader which angle you are talking about. And this is not Physics. The laws and equations of nature must be described in a uniquely manner.

 

[Bob S]

[Bob S]
In the 1950's and 1960's, before the advent of electronic (and pocket) calculators, physicists used trigonometric (sin, tan, cotan, cos) tables (for 90-degree right triangles) in math books to calculate energy and momentum of particles in beamlines; e.g., E sin(theta) = pc, where beta = sin(theta). Also, cos(theta) = sqrt(1-beta2), and E cos(theta) = rest mass. Luis Alvarez I think was one of the earliest to have a HP-35 pocket calculator in the late 1960's, the earliest pocket calculator to have trig, log, and exponential functions.

Dear Bob,
The trigonometric equation you give don't define the ratio E/pc of a uniquely manner.It all depends on which corner of the triangle you take.For example you can also have cos(theta)=E/pc, that means you always have to draw a triangle in order to show to the reader which angle you are talking about. And this is not Physics. The laws and equations of nature must be described in a uniquely manner.

You are correct in that there is more than one way to define what sin(theta) and cos(theta) are. My convention has been to draw m₀c² along the positive x-axis beginning at the origin. I then draw "pc" up vertically along the y direction from the x-axis at point "m₀c²". I then draw "E" from the origin up diagonally to the end of pc. This forms a right triangle, with a slope = sin(theta) = beta. Let's try both methods. Suppose beta = 0.5. What is sqrt(1-beta²). The answer is, using my pocket calculator, 0.866025. Now use trig tables. I find sin(theta)=0.5, and cos(theta)=0.86603. Now let's do it backwards; I look up cos(theta)=0.5, and then sin(theta) = 0.86603!. So it doesn't matter which way I do it. Which way did you do it?

 

[ahdaf]

[Bob S]
In the 1950's and 1960's, before the advent of electronic (and pocket) calculators, physicists used trigonometric (sin, tan, cotan, cos) tables (for 90-degree right triangles) in math books to calculate energy and momentum of particles in beamlines; e.g., E sin(theta) = pc, where beta = sin(theta). Also, cos(theta) = sqrt(1-beta2), and E cos(theta) = rest mass. Luis Alvarez I think was one of the earliest to have a HP-35 pocket calculator in the late 1960's, the earliest pocket calculator to have trig, log, and exponential functions.


You are correct in that there is more than one way to define what sin(theta) and cos(theta) are. My convention has been to draw m₀c² along the positive x-axis beginning at the origin. I then draw "pc" up vertically along the y direction from the x-axis at point "m₀c²". I then draw "E" from the origin up diagonally to the end of pc. This forms a right triangle, with a slope = sin(theta) = beta. Let's try both methods. Suppose beta = 0.5. What is sqrt(1-beta²). The answer is, using my pocket calculator, 0.866025. Now use trig tables. I find sin(theta)=0.5, and cos(theta)=0.86603. Now let's do it backwards; I look up cos(theta)=0.5, and then sin(theta) = 0.86603!. So it doesn't matter which way I do it. Which way did you do it?

I really appreciate the way you say the thing and I liked your first sentence you started with your quote and want you to have a look again at it. You said " there is more than one way to define..." and precisely this shouldn't exist. There should only one way to define thing without any ambiguities.Take this example. If you tell me find the work done by a force F during a displacement L, I know that I have to write W=F.L cos(F,L) and I don't need you to assist me about the angle between F and L. Another example if you have to find the torque of F about a certain point distant r from F. Here you just go and write T=rxF=rFsin(r,F). That is you know already that the angle is between r&R and anyone else.

 

[Bob S]

I really appreciate the way you say the thing and I liked your first sentence you started with your quote and want you to have a look again at it. You said " there is more than one way to define..." and precisely this shouldn't exist. There should only one way to define thing without any ambiguities.Take this example. If you tell me find the work done by a force F during a displacement L, I know that I have to write W=F.L cos(F,L) and I don't need you to assist me about the angle between F and L. Another example if you have to find the torque of F about a certain point distant r from F. Here you just go and write T=rxF=rFsin(r,F). That is you know already that the angle is between r&R and anyone else.

The lack of ambiquity is that If I look up beta in the trig table under either sin (or cos), then sqrt(1-beta²) is found in the trig table labelled cos (or sin). I can also look up sqrt(1-beta²) in sin (or cos), and find beta under cos (or sin). This is not ambiguity. It is sometimes labelled degeneracy (like multiple atomic states having same energy). This has nothing to do with torque.

 

[dbbeery]

I am looking for the origin of the energy triangle. Books refer to Einstein, but I have not found a specific reference. I failed to find it with the help of Google. Has anyone seen a specific reference?

 

[jtbell]

Do you mean the historical origin, i.e. who first used it and when? Or are you looking for a derivation of it?

 

[dbbeery]

Just a reference to the historical origin.