Undergrad E Vector in E=mc2? Scalar Multiplication & Vectors

[andrecoelho]

TL;DR

E=mc2 , E vector?

If the energy is a vector, which as i understand for example, the potential energy , U=mgh, where g is the gravitational force, Then U is the product of scalars and vectors, so its a vector
In that case being E a vector , can it be equal to mc2 (each are scalars). Like mulitplication of scalars are not vectors...

 

[Dale]

Energy is not a vector.

 

[andrecoelho]

But isn't energy the exorthation of force in a body?

 

[Dale]

I have no idea what “exorthation” means. But regardless of what that word means, energy is not a vector. Force is a vector, but that doesn’t mean that everything that has something vaguely to do with force is also a vector.

 

[andrecoelho]

I have no idea what “exorthation” means. But regardless of what that word means, energy is not a vector.

exorthaton -> the effect of applying the force

Can you explain why? Particulary the U=mgh formula

 

[jim mcnamara]

What are the units of energy? A Joule.
https://www.physics.uci.edu/~silverma/units.html
@Dale
When your read this online explanation like this one, you can see where it is possible to get confused.

 

[robphy]

Summary:: E=mc2 , E vector?

If the energy is a vector, which as i understand for example, the potential energy , U=mgh, where g is the gravitational force, Then U is the product of scalars and vectors, so its a vector
In that case being E a vector , can it be equal to mc2 (each are scalars). Like mulitplication of scalars are not vectors...

In U=mgh, the g is not the gravitational force--- m\vec g is the gravitational force (and \vec g is the gravitational field).
More completely, U=-(m\vec g)\cdot \vec h (as minus the work done by the gravitational force).
The dot-product of two vectors is a scalar. So, U is a scalar.
In fact, the gravitational potential gy is a scalar field
(so minus its gradient is the gravitational field)
and so the gravitational potential energy mgy is a scalar.

Note that ordinary kinetic energy (from the work-energy theorem) is a scalar K=\frac{1}{2}mv^2.

Energy is a scalar.

 

[PeroK]

exorthaton -> the effect of applying the force

Can you explain why? Particulary the U=mgh formula

The $g$ in that formula is the magnitude of the gravitational force: $g = |\vec g|$. More generally, we have $$U = -\frac{GMm}{r}$$
In any case, potential energy is not a vector; nor is any form of energy. Kinetic energy is $\frac 1 2 mv^2$, where again $v = |\vec v|$ is the magnitude of the velocity. So, KE is also a not a vector.

As you've posted this in relativity forum, the energy of a particle is a component of the energy-momentum four vector: $$\mathbf{p} = (E, p_x, p_y, p_z) = (E, \vec p)$$ In that sense, energy is a component of a four-vector.

 

[vanhees71]

Keeping $c$ it's
$$(p^{\mu})=(E/c,\vec{p}).$$
One should also use invariant masses and only invariant masses in relativity, and thus the relation between energy and momentum of a particle reads
$$p_{\mu} p^{\mu}=\frac{E^2}{c^2}+\vec{p}^2=m^2 c^2$$
or
$$E=c \sqrt{(m c)^2+\vec{p}^2}.$$
In terms of the coordinate velocity $\vec{v}=\vec{p} c^2/E$ it reads
$$E=\frac{m c^2}{\sqrt{1-\vec{v}^2/c^2}}.$$
The famous formula by Einstein should read (BTW also according to Einstein himself!)
$$E_0=m c^2,$$
where $E_0$ is the energy as measured in the rest frame of the particle.

 

[Dale]

Can you explain why? Particulary the U=mgh formula

In general a force and its associated potential energy have the relationship: $\vec F = -\nabla U$. Since $\nabla$ is an operator that converts a scalar into a vector it is clear that $U$ must always be a scalar.

In the case of a uniform gravitational field we have $$\vec F = m \vec g = -\nabla (mgh) = -\nabla U$$ where on the left hand side $\vec g$ is a vector pointing downwards and on the right hand side $g$ is the magnitude of that vector.

 

[andrecoelho]

What is the relevance of attributing a unit to a expression? does that mandate that the quantity is vectorial?!
what is the relation between a joule and the fact that its possible a vector?
And i still don't get it, h is not a vector , is distance
" If you accept that energy, by definition, is a scalar function," , i can't just assume that something just by definition, its not an axiom. If the math is incorrect, its also incorrect in terms of physics, because physics use exclusive maath
is the h in the formula a vector (lisnt length a scalar)??
Energy can be transferred one side to another, hence could be a vector the problem remains,
is energy released to every part of a body, and each part is the same? or is there a parts of the body with more energy
than others? Just because we don't know what happens to energy after being transferred (for example The force used to store energy in a rubber band has a certain direction,
that doesn't mean its not a vector, so one thing is sure, energy can be trasferred (which applies being vector) . If you don't know where the energy goes,
how can you know its not a vector

Does the energy gets released equally?
or the whole body has energy (where other parts dont). Does energy already remeains at a body and energy gets added
beign released means going into a direction or several

 

[hutchphd]

I prescribe a little Feynman and get back to us after:
https://www.feynmanlectures.caltech.edu/I_04.html

 

[Dale]

What is the relevance of attributing a unit to a expression? does that mandate that the quantity is vectorial?!

No, assigning units in no way makes the quantity a vector. I am not sure what would possibly lead you to believe that.

i can't just assume that something just by definition

Why not? Definitions are true by definition.

Energy can be transferred one side to another, hence could be a vector the problem remains,

Actually, no. What you are thinking of is energy flux, not energy. Energy flux density is a vector.

If you don't know where the energy goes,
how can you know its not a vector

We know that it is not a vector because it is defined to not be a vector. Since it is a definition it is true by definition.

energy can be trasferred (which applies being vector)

This is nonsense. Money can also be transferred. Money is not a vector. Transferred applies more to conserved quantities than to vector quantities.

There really is no way around this and absolutely no ambiguity on this topic whatsoever. Energy is not a vector.

Since there is literally nothing more to say on this topic and since you seem disinclined to actually learn anything here, this thread is closed.