![](https://www.scienceforums.net/uploads/set_resources_1/84c1e40ea0e759e3f1505eb1788ddf3c_pattern.png)
![](https://www.scienceforums.net/uploads/set_resources_1/84c1e40ea0e759e3f1505eb1788ddf3c_default_photo.png)
PaulS1950
Senior Members-
Posts
206 -
Joined
-
Last visited
About PaulS1950
- Birthday 09/10/1950
Profile Information
-
Location
Central Puget Sound
-
Occupation
retired
Retained
- Baryon
PaulS1950's Achievements
![Rank: Atom (5/13) Atom](https://www.scienceforums.net/uploads/set_resources_1/84c1e40ea0e759e3f1505eb1788ddf3c_default_rank.png)
Atom (5/13)
23
Reputation
-
This is an assumption made about untested data. No one has determined the properties of the Higgs Boson or the higgs field. No one knows what effects the field may or may not have - it is all conjecture. Paul
-
Since the photon can't exist at rest you are right; the REST mass of a photon is zero. If you don't exist your mass is also zero. A photon, in its natural frame, has mass because of its velocity (at least). Paul
-
It is mass and energy that affects the spacetime not gravity. Gravity is what we call the effect of bent spacetime. Paul
-
[math] E=mc^2 + pc[/math] from [math] E^2 = m^2c^4 + p^2c^2[/math] by finding the square root of both sides of the equation. so [math] m=E/(c^2+pc) [/math] Since the photon exists only at the speed of light it is as improper to find the "rest" mass as it is to find your personal mass at c...... Both are meaningless. The only mass for a photon that has any meaning is the relativistic mass. Paul
-
If the momentum is E/c then the mass is[math] E/c^2.[/math]
-
So you start with the assumption that the mass is zero in an equation to find the mass? How absurd is that? If I assume the mass is [math] 10 ^-1000000000000[/math] then I will "prove" that it does have mass. Start with the values of momentum and / or energy of a photon within its reference frame and calculate mass only from that data.
-
How much mass would you have if you did not exist? How much mass would you have at the velocity C? A photon has mass because it is at velocity C. Taking it out of that reference frame is the same as my first question in this post. The second question is as meaningless as an at rest photon. Neither can happen.
-
[math]E^2= m^2c^4+\mathbf{p}^2c^2[/math] reduces to [math]E= mc^2 + \mathbf{p}c[/math] not E=pc you seem to be missing part of the formula?
-
The frequency adds mass? I believe that mass times velocity is momentum.
-
We all know that you can't use zero when multiplying or dividing - it isn't proper math because zero is valueless. If p=0 then it has no meaning. cp = 0 but you don't get zero because p = x>0. If you multiply infinity by zero you still get zero (no infinity's) What is the value of p in E=pc? it has to be >0
-
The only reason a photon has energy or momentum is because of its velocity. "At rest" there is nothing - the photon can't exist. Therefore it is meaningless to have a discussion about the mass of an object that doesn't exist. Of course it has zero mass - it isn't there! Since its velocity is a function of its existence then we have to calculate its, energy, momentum and MASS in that reference frame. What is the mass of a photon traveling at C? What is the mass of a billion photons travelling at C? The momentum is a billion times higher than one... If a photon has energy and momentum then it MUST have mass (even if it is a result of its velocity which is a function of its existence). If M=0 then MC^2 =0 it cannot have kinetic energy (it does) and MC=0 it cannot have momentum (it does), it follows logically and mathmatically that a photon must have mass. M cannot be zero. Paul
-
Using a perfectly balanced beam which pivots at its center on a frictionless pivot, one side of the beam resting on the scale and the other end attached to the balloon filled with helium your scale will measure the net lifting force of the balloon. The net lifting force will be equal to the pressure density of the atmoshere (under test conditions) minus the pressure density of the helium minus the mass of the balloon. Paul
-
We all agree that light has momentum. (m for notation sake only - Mass will use M) We all agree that it has a velocity of C. The mass of an object is the momentum divided by its velocity. (in units weight time interval) (foot pound seconds for projectiles) This is converted to slugs or newtons by way of reducing weight to mass. Without assuming a mass what is the momentum of a photon? now for the sake of measuring the very small let us use a scale that that multiplies the number (not the actual momentum just the number used to define it) by 1000000. We can call this scale nx's and note that a proton would have many millions of nx units as its mass. so, take the momentum of a photon, multiply by 1000000 and divide by C. What number do you end up with?
-
is it possible for a spring powered automobile?
PaulS1950 replied to cameron marical's topic in Engineering
Not unless you find a way to bend or break the laws of physics. You can run an engine at a "most efficient for mileage" range of RPM but you cannot use it to efficiently wind a spring and then run a car with it. The engine has to put out more torque than the spring produces, otherwise it wouldn't be able to wind it up. The the car running on the spring would have less power than it took to wind it so more loss. Short version is that it always take more power to operate a system than it produces. Running alternaters on the wheels of an electric car doesn't work because you ave to have enough power to run the alternators and move the car. alternators are 90% efficient (they produce 90% of the watts that it takes to spin them) an electric motor only produces 90(~) % of the power that is fed into it. If you use a gas engine to run an alternator to run an electric motor you have the losses of the gas engine plus the 19% loss in the electrical system. If you are running a controller for the electric motor then it also consumes power. At the end of the day, month, year, decade or century nothing is free. There are always losses. you have to use more energy to produce motion than the motion can provide in energy.